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In L–C–R series AC circuit voltage amplitude across L, C and R are 10V, 6Vand 3V respectively. At an instant voltage across the resistance is 3√3 /2 volt and source voltage is decreasing then what can we say about the current in the circuit at that instant.(increasing or decreasing) explain

In L–C–R series AC circuit voltage amplitude across L, C and R are 10V, 6Vand 3V respectively. At an instant voltage across the resistance is 3√3 /2 volt and source voltage is decreasing then what can we say about the current in the circuit at that instant.(increasing or decreasing) explain

Grade:12th pass

1 Answers

Vikas TU
14149 Points
5 years ago
Current in the L-C-R circuit,
i = Vsource/Znet = > (10 + 6 + 3)/root((wl)^2 + (1/wc)^2 + R^2) = > 19/root((wl)^2 + (1/wc)^2 + R^2)
i’ = Vsource/Znet => (10 + 6 + 3root3/2)/Znet = >(16 + 3root3/2)/root((wl)^2 + (1/wc)^2 + R^2)
=> 18.59/root((wl)^2 + (1/wc)^2 + R^2)
Thus source voltage decreases and net impedence reamins same.
the current in circuit will too decrease.
as it is directly proprotional to vilatage source.

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